How do I simplify this equation? I am using it to find the tangent equation to a circunference which is x^2+y^2-8x+3=0

[lysss]

Hello.

I am stuck where I need to simplify.

The problem asks to find the equation of the tangent to the circumference x² +y² -8x+3=0 with points (6,3)

I started by using
Y-Y1=m(X-X1)
So,
Y-3 = m (X-6)
Y= mx-6m+3

I replaced the value of Y in the original equation so it's

x² +y² -8x+3=0
x² +(mx-6m+3)²-8x+3=0
And then I did this:

(m²+1)x² +36m²+9-8x+3=0

And I do not know if that is right and don't know what's next...

Thank you for your help in advance

 

[Deleted member 4993]

Hello.

I am stuck where I need to simplify.

The problem asks to find the equation of the tangent to the circumference x² +y² -8x+3=0 with points (6,3)

I started by using
Y-Y1=m(X-X1)
So,

Y-3 = m (X-6)​

Y= mx-6m+3

I replaced the value of Y in the original equation so it's

x² +y² -8x+3=0
x² +(mx-6m+3)²-8x+3=0
And then I did this:

(m²+1)x² +36m²+9-8x+3=0

And I do not know if that is right and don't know what's next...

Thank you for your help in advance

The best way to calculate the slope of the tangent-to-circle is to use the fact:

Tangent line, at a given point on the circle, is perpendicular to the radial line at that point.​

The circle is → (x- 4)² + y² = (√13)² → the center of the circle is at (4,0) ............. corrected (Jomo I am off to the corner)

You know the radial line has to go through (4,0) and (6,3).

What is the slope of that radial line?

This radial line is perpendicular to the tangent line at (6,3).

What is the slope of that tangent line?

and continue......

 

[lex]

You haven't used the fact that this line is tangent to the circle at (6,3), just that it meets the circle at (6,3)
Oh I see Subhotosh Khan has answered. I'll leave it to him!

 

[pka]

The circle \(x^2-8x+y^2+3=0\) in standard form is \((x-4)^2+y^2=13\).
On the circle the slope at each point is \(y'=\dfrac{-(x-4)}{y}\)

 

[Harry_the_cat]

If you want to use calculus, you could find dy/dx using implicit differentiation. This will give you the "formula" for slope at any given point on the circle (that pka has given in post #4).
If you don't know how to do that, then follow Subhotosh's advice. Don't be afraid to draw a diagram to help you think it through!

 

[lex]

Just a note, connecting the two methods: you can of course get pka's formula [MATH]\frac{dy}{dx}=-\frac{(x-a)}{(y-b)}[/MATH] by Subhotosh Khan's radial method too.